KENDRIYA VIDYALAYA SANGATHAN, Jaipur Region

Pre-Board Examination: 2024-25

Class: 12th Subject: Biology (044) SET No._____
M.M: 70 Marking Scheme Time: 3 Hours

Maximum Marks: 70 Time: 3 hours

Q.N SECTION MAR

O. A KS
1 (C) Two polar nuclei+1Male gamete nucleus 1
2 (C) the flower is cleistogamous. 1

3 (a) Least active form of virus 1

4 (B) Life comes only from pre-existing life 1

5 (B) Down’s syndrome Turner syndrome 1
6 (A) Lactobacillus and Yeast 1

7 (b) Primary consumer 1

(A) Performing a cross between tall pea plant with unknown genotype with a dwarf
8 1
pea plant
9 (D) It remains active at high temperature
10 (C) Inheritance of an autosomal recessive trait and pleiotropy 1
11 (C) It contains alpha-lactalbumin like mother milk 1
12 (A) move away from equator to poles 1
13 (A) Both A and R are true and R is the correct explanation of A. 1
14 (C) A is true but R is false. 1
15 (C) A is true but R is false. 1
16 (C) A is true but R is false 1
SECTION
B
17 advantages of emasculation – prevent autogamy in only plants bearing bisexual ½+½
flowers. advantage of bagging – prevent contamination of stigma by unwanted ½+½
pollens in both plants bearing unisexual and bisexual flowers.
OR

18 a. The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) 1+ ½
at the sixth position of the beta globin chain of the haemoglobin molecule due to +½
the single base substitution at the sixth codon of the beta globin gene from GAG
to GUG. The mutant haemoglobin molecule undergoes
polymerisation under low oxygen tension causing the change in the shape of the
RBC from biconcave
disc to elongated sickle like structure
b. genotype: HbS HbS
it is autosomal recessive trait can inherit from Mohan to his son when mother is
carrier for this trait.
19 (i) Protozoan disease- malaria, scientific name of the causative organism- ½+½
Plasmodium vivax/falciparum/malaria ½+½
Helminthes disease- filariasis, scientific name of the causative organism-
Wuchereria bancrofti/malayi
20 (a) Immunity will become weak and person will be prone to diseases. ½+½
(b) AIDS, HIV ½+½
OR

(a) Bacillus thuringiensis produce an antitoxin which when ingested and ½+

released in the gut of the larva of insect pest. Disrupt its guts lining and the ½
caterpillar are killed. +1

(b) Infectious diseases treatment became easy.

21 (A) significance of TetR – selectable marker used to identify and elimination of ½+½
non- transformants and permitting the growth of only transformants +1
BamH I: cloning site to introduce desired gene into vector
(B) No, it does not contain gene for beta-galactosidase which act as selectable
marker for blue-white selection of recombinants from non-recombinants.

OR
1) calcium treatment
2) Heat and cold treatment 1/2X4=2
3) Microinjection
4)Gene gun.

SECTION C
22 a) hormone: FSH ; source: Pituitary gland 1+1+
b) C = mature graffian follicle and cavity present in C but absent in A is 1
Antrum.
c) D = corpus luteum and role of the hormone (progesterone) produced by it is
to maintain endometrium or pregnancy hormone
23 (a) Due to activation of oncogenes or change of proto oncogenes into oncogenes. 1+2

(b) Suitable answer

24 a) ‘X’ =Morula ; ‘Y’= Blastocyst ‘Z’= Trophoblast ½x

6
b) ‘Q’ = oral pills; chemical composition = progestogens or progestogen- =3

estrogen combination

25 (a) DNA can be cut at specific sites by using enzymes restriction 3

endonucleases. These are also known as molecular scissors.
(b) Various processes involved in recombinant DNA technology are :
(i) Isolation of the genetic material (DNA),
(ii) Cutting of DNA at Specific Locations,
(iii) Amplification of Gene of Interest using PCR,
(iv) Insertion of Recombinant DNA into the Host Cell/Organism,
(v) Obtaining the FOREIGN Gene Product.
(c) The cut DNA can be joined together using DNA ligases Enzymes.
26 A=Papaver somniferum, B= depressant and slows down body functions; C= ½x6
Cannabinoids D= effects on cardiovascular system; E= Erythroxylum coca and F=
stimulate CNS, Euphoria
27 a) DNA fragments are negatively charged molecules they can be separated 1+1+
by forcing them to move towards the anode under an electric field. 1

b) most commonly used matrix is agarose which is a natural polymer extracted

from sea weeds.
c) The separated DNA fragments can be visualised only after staining the DNA
with a compound known as ethidium bromide followed by exposure to UV
radiation appeared as dark orange bands.

28 Gross primary productivity of an ecosystem is the rate of production of organic 1+1+

matter during photosynthesis in a given area per unit time. A considerable 1
amount of GPP is utilised by plants in respiration.
Gross primary productivity minus respiration losses (R), is the net primary
productivity (NPP). GPP – R = NPP ; Net primary productivity is the
available biomass for the consumption to heterotrophs (herbiviores and
decomposers).
Secondary productivity is defined as the rate of formation of new organic matter by
 consumers.

SECTION D
Q. No. 29 and 30 are case-based questions. Each question has 3 subparts
with internal choice in one subpart.
29 (a) - Stabilizing selection occurs when the intermediate traits of a population, the 2+
intermediate wing colour in this case, are favoured over the extreme traits, such 2
as wings with extremely bright or dull colours.
(b) - The darker environment aids camouflage for dull-winged butterflies, reducing
predator visibility. Over time, the population could shift towards duller-winged
individuals for increased survival and reproduction in the altered habitat.
OR
(a) A - No microbial growth would be observed since it was sterile and sealed. –
B - Microbial growth could potentially occur due to airborne contaminants reaching the
nutrient media. –
C - Microbial growth from the initially spread culture would likely be observed, as the
sealed environment would prevent external contaminants from entering.
(b) [0.5 mark each for each of the following points] - It disproves the theory of
spontaneous generation. - The lack of growth on the sterile, sealed plate (A)
contradicts the idea of life spontaneously forming. (c) Louis Pasteur's swan-necked
30 (a) Adaptive radiation 4
(b) This diversification allows the fish in the lake to - exploit different food sources - exploit
different habitats - reduce competition - maximizing their chances of survival - increase
beneficial traits and specialization - creates a more stable ecosystem
[Accept any other valid answer]

SECTION E

31 a. Pollen pistil interaction- 3+1

I.chemical components of the pollen interacting with those of the pistil. +1
II.The pistil accepts the right type pollen and promotes fertilisation.
III.Pollen grain forms a pollen tube which emerges through the germ pore.
IV. The contents of the pollen grain (nucleus of vegetative cell and two male
gametes) move into the tube.
V. Pollen tube grows through the tissues of the stigma and style and finally the
ovule through the
micropyle (chemotropically) and then enters one of the synergids through the
filiform apparatus
VI. Tip of pollen tube bursts and the two sperms enter the embryo sac.

b. Syngamy= fusion of male gamete with egg cell and triple fusion= fusion of
second male gamete with two polar nuclei
c. The PEN undergoes repeated division to give rise to free nuclei to form
coconut water(liquid endosperm) and Cell wall formation starts subsequently
to form coconut kernel(cellular endosperm)
OR
a. Sporogenous tissue ½
b. ½ x 5= 2 ½
 c. Diagram

32 Hershey and Chase experiment: ½ x

Experimental Material: 10=
1. Bacteriophages (viruses that infect bacteria)
2. Bacteria E. coli growing in culture medium
5
Steps of Experiment:
 Infection
 Blending
 Centrifugation
Conclusion: DNA is therefore the genetic material that is passed from virus to
bacteria
 2+1
OR +2=
cross between homozygous dominant for yellow and round seeds 5
(RRYY) with homozygous recessive (rryy) up-to F2 progeny showing
phenotypic ratio of 9: 3: 3: 1.

½ x 4 =2
b. Statement of law of independent assortment: Mendel's law of independent
assortment states that the alleles of two (or more) different genes get sorted
 into gametes independently of one another. In other words, the allele a
gamete receives for one gene does not influence the allele received for
another gene. 1
c. Write the possible genotypes for the progeny for such a cross having
(i) Yellow and wrinkled seeds – Yyrr, YYrr ½ x 2=1
Green and round seeds – yyRR, yyRr ½ x 2=1

33 (a) dung + water ½ x 2 1+1

(b) methanogens, and anaerobic condition ½ x 2 +1+
(c) methane along with CO2 and H2 ½ x 2 2=5
(d) monera- bacteria and cyanobacteria -azotobacte or rhizobium/nostoc= nitrogen
fixation
½x2
Fungi- mycorrhiza- Absorbs phosphorus from soil and passes to plant ½ x 2
OR
1x5
a. Macrophage and T-lymphocytes ½ x 2 =5
b. B – viral RNA and enzyme ‘C’ - reverse transcriptase. ½x2
c. Retrovirus: A retrovirus is a virus that uses RNA as its genomic material.
Upon infection with a retrovirus, a cell converts the retroviral RNA into DNA. ½ x
2
d. Lysis of host cell 1
Decrease in immunity /immunodeficiency 1